I was going through Penn State's online notes and noticed this expression:

$ v^2 = \frac{1}{n} \sum_{i=1}^n (y_i - \bar{y})^2$

In the line below it they stated that the $E[v^2] = (1 - \frac{1}{n})\sigma^2$. I was wondering how would you get that?

Would it be wrong for me to say that since $\sum_{i=1}^n (y_i - \bar{y})^2 = (n-1)s^2$ the $E[v^2] = E[\frac{(n-1)s^2}{n}] = (1 - \frac{1}{n})\sigma^2$ ? But this line of reasoning forces me to assume that $E[s^2] = \sigma^2$ and I don't even know why that's true.

  • $\begingroup$ Notes for what? What is the exact definition of $s^2$? It looks a bit like something related to finding an unbiased estimator for the sample variance. But perhaps a bit of clarification would be helpful. $\endgroup$ Oct 27, 2013 at 23:02
  • $\begingroup$ nevermind i figured it out $\endgroup$
    – Person
    Oct 27, 2013 at 23:24
  • $\begingroup$ Good job! _____ :) $\endgroup$ Oct 27, 2013 at 23:28
  • $\begingroup$ Incidentally, the relevant Wikipedia page has it... but uses appalling notation ($\sigma^2_y$ for a sample quantity?! So much for avoiding misunderstanding by tossing convention to the winds.) $\endgroup$
    – Glen_b
    Oct 28, 2013 at 1:12

1 Answer 1


Ok, though I'm not sure about what is the course, or what is your $s^2$ exactly, the course notes seem to use $v^2$ to denote sample variance.

The proof (see bit on sample variance) of the line is not hard, but I guess "hard" is relative. I remember, a lot of people struggling with this one in the undergrad. The key is not to forget, after writing out the second power of the term in brackets and taking the expected value operator inside the sum, you have to use $E[X^2] = V[X^2] + E[X]^2$ to substitute all the terms $E[y_i^2]$, the rest is re-arranging the sums.


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